cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Wolfram Hüttermann

Wolfram Hüttermann's wiki page.

Wolfram Hüttermann has authored 3 sequences.

A289690 Least k such that there are exactly n perfect powers between 10k and 10k + 10.

Original entry on oeis.org

5, 1, 2, 12, 0
Offset: 0

Views

Author

Wolfram Hüttermann, Jul 09 2017

Keywords

Comments

I do not know if a(5) exists. If it does, the numbers 10k+1, 10k+3, 10k+5, 10k+7, 10k+9 will be perfect powers. But those numbers are very scarce.
Further, a(6), ..., a(10) cannot exist because of the Mihailescu theorem, as the only adjoining perfect powers are 8 and 9.
a(5) is extremely unlikely to exist; if it does, it is larger than 10^70. - Charles R Greathouse IV, Jul 21 2017

Examples

			If n=2, then there are 2 power numbers between 20 and 30: 25 and 27, and this is the least k with this property.

Crossrefs

Cf. A001597, A097056.

Programs

  • PARI
    a(n)=my(k=0); while(sum(j=10*k+1, 10*k+9, (j==1) || ispower(j)) !=n, k++); k; \\ Michel Marcus, Jul 20 2017

A282059 Numbers k where there are 8 primes between 10*k and 10*k + 30.

Original entry on oeis.org

1, 8879, 28472, 85571, 114677, 656099, 1576009, 2565844, 6915653, 7426625, 9362599, 18240349, 21803372, 22644952, 26167277, 30254276, 66197230, 91093591, 96466961, 104209078, 107132278, 117022186, 134030186, 139402516, 140053322, 142247591, 145927027
Offset: 1

Views

Author

Wolfram Hüttermann, Feb 05 2017

Keywords

Comments

8 primes are the maximum in such an interval. Dickson's conjecture would indicate that there are infinitely many k. It is easy to prove that k = 21*m + r with r in {1, 17}.

Examples

			There are eight primes between 88790 and 88820: 88793, 88799, 88801, 88807, 88811, 88813, 88817, 88819. Therefore 8879 is in the sequence.

Crossrefs

Cf. A078854, A007530 (prime quadruplets).
For the Dickson conjecture see A020497.

Extensions

a(18)-a(27) from Giovanni Resta, Mar 29 2017

A279862 a(n) = the smallest integer k where there are exactly n primes between 10k and 10k+100.

Original entry on oeis.org

37027, 15590, 3562, 3561, 1881, 1856, 735, 588, 132, 131, 188, 111, 89, 47, 44, 32, 20, 11, 9, 8, 5, 3, 2
Offset: 0

Views

Author

Wolfram Hüttermann, Dec 21 2016

Keywords

Comments

a(25) = 0, and a(24) cannot exist. The same is true with a(k) and k>25. From A020497, we see that a range of 101 numbers is required to find 24 primes. It is an open question if a(23) exists.
From Robert Israel, Jan 18 2017: (Start)
Dickson's conjecture implies that a(23) does exist.
Let Q = 27926129625869590, and R = 614889782588491410 the product of all primes < 50.
Then for any k, the 23 numbers Q+i+k*R for i = 1, 3, 7, 9, 13, 19, 21, 27, 31, 33, 37, 43, 49, 51, 57, 63, 69, 73, 79, 87, 91, 97, 99 have no prime divisors < 50.
Dickson's conjecture would indicate that there are infinitely many k for which these numbers are all prime, and thus there are 23 primes between Q+k*R and Q+k*R+100. (End)
Heuristics suggest a(23) exists (see above) and has between 20 and 30 digits. There are 192 residue classes mod 23# = 223092870 in which a(23) might fall, all of which are 11 mod 30 and either 3 or 4 mod 7. - Charles R Greathouse IV, Jul 12 2017

Examples

			For n = 1 there is only one prime between 155900 and 156000: 155921.

Crossrefs

Cf. A186311.
For the Dickson conjecture, see A020497.

Programs

  • Maple
    for n from 1 to 10^5 do
  T[n]:= nops(select(isprime, [$10*n+1 ..10*n+9]))
od:
for k from 1 to 10^5-10 do
   v:= add(T[k+j],j=0..9):
   if not assigned(A[v]) then A[v]:= k fi
od:
seq(A[n],n=0..22); # Robert Israel, Jul 12 2017
  • Mathematica
    Function[s, -1 + Flatten@ Table[FirstPosition[s, n] /. k_ /; MissingQ@ k -> 0, {n, 0, Max@ s}]]@ Table[Count[Range[10 k, 10 k + 100], ?PrimeQ], {k, 0, 10^5}] (* _Michael De Vlieger, Jul 12 2017; program writes "-1" for a(23) and a(24). *)
  • PARI
    a(n) = my(k=0); while(1, if(primepi(10*k+100)-primepi(10*k)==n, return(k)); k++) \\ Felix Fröhlich, Jul 12 2017
  • PARI
    a(n)=my(k); while(sum(p=10*k+1,10*k+99,isprime(p))!=n, k++); k \\ Charles R Greathouse IV, Jul 12 2017

Extensions

Name clarified by FUNG Cheok Yin, Jul 12 2017

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.